Đáp án:
$\min y = -\sqrt{5 + 2\sqrt2};\, \max y = \sqrt{5 + 2\sqrt2}$
Giải thích các bước giải:
$y =2\cos x - \sin\left(x-\dfrac{\pi}{4}\right)$
$\to y = 2\cos x - \left(\dfrac{\sqrt2}{2}\sin x - \dfrac{\sqrt2}{2}\cos x\right)$
$\to y = \left(2 +\dfrac{\sqrt2}{2}\right)\cos x - \dfrac{\sqrt2}{2}\sin x$
Ta có:
$y^2 \leq \left(2 +\dfrac{\sqrt2}{2}\right)^2 + \left(\dfrac{\sqrt2}{2}\right)^2$
$\to y^2 \leq 5 + 2\sqrt2$
$\to -\sqrt{5 + 2\sqrt2}\leq y \leq \sqrt{5 +2\sqrt2}$
Vậy $\min y = -\sqrt{5 + 2\sqrt2};\, \max y = \sqrt{5 + 2\sqrt2}$
